Today are going to be talking about how to solve a boundary value problem.
And in this particular problem we've been given the second order homogenous differential equation why
double prime or the second derivative of Y plus for Y is equal to zero.
We've also been given two initial conditions y of zero equals 5 and y of Pi over four equals three.
So in order to solve this problem the first thing we're going to need to do is solve for the roots of
our second order homogenous differential equation.
The way that we're going to do that is first by converting this equation from y variables to our variables.
And the way that we're going to do that is by using this silly little trick that I have where I count
the hashmarks.
These prime marks here on each Y variable to determine what the exponent will be on the R variable.
So for example this first term here is why double prime.
It has 2 hashmarks right here.
What that's going to convert to therefore is r squared to hashmarks the new exponent is 2.
So why double prime becomes R-squared then we're going to have plus four.
And now to convert this Y variable into an R variable again we're going to count the hashmarks.
There are none of them.
There is zero.
So we're going to have R to the zero in our new equation.
Well ours of the 0 is just one.
So we can just leave this as for four times one is just for.
We'll leave it as four.
So we get our squared plus four is equal to zero.
And once we've converted our equation from y variables to our variables we want to go ahead and solve
for our own when we solve for our what we're going to get are the roots or the solutions to this second
order differential equation.
So we'll go ahead and subtract 4 from both sides and we'll get our squared is equal to negative 4.
If we take the square root of both sides what we'll end up with on the left hand side is our on the
right hand side will have positive or negative square root of negative 4.
Now as you may know the way that we're going to take the square root of this negative number is to involve
the imaginary number I.
What we're going to do is we're going to call this r equals positive or negative square root of four
times negative 1.
So what we've done here is we've separated the negative from this value here for we haven't changed
the value because four times negative 1 is negative 4.
But we've gotten this negative one here by itself like this underneath the square to what that's going
to allow us to do is take the square root of 4.
That's just 2.
So we're going to get our equals positive or negative 2 times the square root of negative one.
Well the square root of negative one exactly is I.
We can cancel that out and call this I.
It's the same thing I is the square root of negative 1.
So instead of having a squirt of negative 1 we're going to have AI and so we're going to say R is equal
to positive or negative 2.
I and that's how we simplified to find the roots of this second order differential equation.
Now what this should tell us remember that when we're dealing with second order differential equations
we're going to get one of three types of roots we're going to get either distinct real roots equal real
roots or complex conjugate roots.
I have other videos about distinct real roots and equal real roots in this particular case.
We have complex conjugate roots.
Whenever we get roots of the form outfought plus or minus beta times I those are complex conjugate roots.
In our case the roots are our equals zero plus two II and zero minus two I we've just broken out this
positive negative sign here.
And of course that ends up just simplifying to two.
Or negative too.
Those are complex conjugate roots.
But what this tells us notice if we put this zero here in from these two roots here zero plus or minus
two.
That's an exactly the same form as this Alpha plus or minus beta I.
And what that tells us is that in our problem Alpha is equal to zero and Beta is equal to 2.
We just literally equate this template here with our roots and we can see that Alpha is zero.
And that too is equal to beta.
So we get a value for alpha and beta and then we can use the formula for the general solution when we're
dealing with complex conjugate roots which I've written here as y of X..
So remember them you use a different formula for the general solution.
When you're dealing with distinct real roots or equal real roots when it's complex conjugate roots like
we have this is the formula we're going to use.
So our general solution is why of X equals e to the Alpha x y Alpha is zero.
So we're going to get to the zero x or just e to the zero.
Each of the zero is just one so we actually don't even need to write that at all.
What we're left with then is CSUB one cosign of beta X beta is 2.
So we have cosign of two x plus C sub to sign of 2 x.
So this is equation of our general solution now to solve the boundary value problem what we need to
do is start plugging in our initial conditions.
So I will take this first initial condition wives observe equals five.
And what we'll do is we'll set this whole thing equal to 5 and we'll plug zero in to the right hand
side for x.
What we're trying to do with these initial conditions is solve for the constants.
See someone in CSUB too.
One interesting thing to note about boundary value problems.
They're not always solvable you can't always find a solution like you can with initial value problems.
That's the thing that makes them different from one another boundary value problems versus initial value
problems.
We're not guaranteed to find a solution here but we can sure as heck try.
So we're going to plug in zero on the right hand side here and we're going to get CSUB one times cosign
of two times zero so we're going to get cosign of zero so cosign here of zero plus CSUB to again here
sign of zero.
Now we know that cosign of zero is equal to 1.
So we'll just get CSUB one times one or see someone sign of zero is equal to zero.
So zero time CSUB two is still zero.
That whole thing's going to go away.
And what we see here is that we have CSUB one is equal to five.
So we've already solved for one of our constants.
Now if we plug in our second initial condition to our general solution hopefully we'll be able to find
CSUB too.
So when we do that we'll say three equals CSUB one times cosign of 2 times Piver for two terms power
for us just over 2.
So we'll get cosign of Pi over to plus CSUB 2 times sign of Pi over 2.
So get three equals here first see someone we know see someone is 5 so we can go ahead and plug that
in five cosign of Pi over to is zero so we get zero there plus CSUB to sign of Pi over to is equal to
1.
So we get that this is going to go away and you can see with it what we're left with is three equals
C sub 2.
And we've solved for our second constant CSUB 2.
So that's the goal to solve for both constants.
When we do we want to go ahead and plug both of them back into the general solution which we found earlier
the general solution here.
So we're going to get for our final answer is formula or an equation for the general solution Y of x
is equal to.
Here we have CSUB ones who are going to get five times cosign of 2 x plus CSUB 2 which we know is three
so three times sign up to X and that's it.
This is our final answer for the boundary value problem.